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If n is a positive integer, the sum of the integers from 1 to n, inclu

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If n is a positive integer, the sum of the integers from 1 to n, inclu  [#permalink]

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New post Updated on: 30 Oct 2018, 01:12
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If n is a positive integer, the sum of the integers from 1 to n, inclusive, equals \(\frac{(n(n+1))}{2}\) Which of the following equals the sum of the integers from 1 to 2n, inclusive?


A. \(n(n+1)\)

B. \(\frac{(n(2n+1))}{2}\)

C. \(n(2n+1)\)

D. \(2n(n+1)\)

E. \(2n(2n+1)\)

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Originally posted by pushpitkc on 30 Oct 2018, 01:08.
Last edited by Bunuel on 30 Oct 2018, 01:12, edited 1 time in total.
Edited the question.
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Re: If n is a positive integer, the sum of the integers from 1 to n, inclu  [#permalink]

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New post 30 Oct 2018, 01:29
pushpitkc wrote:
If n is a positive integer, the sum of the integers from 1 to n, inclusive, equals \(\frac{(n(n+1))}{2}\) Which of the following equals the sum of the integers from 1 to 2n, inclusive?


A. \(n(n+1)\)

B. \(\frac{(n(2n+1))}{2}\)

C. \(n(2n+1)\)

D. \(2n(n+1)\)

E. \(2n(2n+1)\)


sum of the integers from 1 to n, inclusive, equals \(\frac{(n(n+1))}{2}\)

Replacing \(n\) by \(2n\)

sum of the integers from 1 to \(2n\), inclusive, equals \(\frac{(2n(2n+1))}{2} = n(2n+1)\)

Answer: Option C
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Re: If n is a positive integer, the sum of the integers from 1 to n, inclu   [#permalink] 30 Oct 2018, 01:29
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